Question

8. What is the chance of throwing a sum greater thar
or equal to 7 in a throw of 2 dice?​

Answer 1

Throw of a single die can give 6 probable outcomes; namely, 1, 2, 3, 4, 5 & 6.

So, throw of two dice can give you 6 * 6 = 36 probable outcomes.

Sequences leading to getting at least 7 = (1, 6), (2,5), (2,6), (3, 4), (3, 5), (3, 6), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) & (6, 6).

Therefore, probable outcomes for getting at least 7 = 1 + 2 + 3 + 4 + 5 + 6 = 21

So, the probability of getting at least 7 from throwing of two dice = 21 / 36 = 7 / 12